Theorem: Let $m$ be a positive integer. Then $H_m(\mathbb{R}^n)=\{ u \in D'(\mathbb{R}^n): D^{\alpha} u \in L^2(\mathbb{R}^n), |\alpha| \leq m\}$
$\to ||u||_{H_m}^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}<+\infty$
Furthermore, $H_m(\mathbb{R}^n)$ is the completion of $C_C^{\infty}(\mathbb{R}^n)$ ad for the metric $( \star) ||u||_m=\left( \sum_{|\alpha| \leq m} \int |D^{\alpha}u|^2\right)^{\frac{1}{2}} $
Proof: There are constants $c_m$ such that $$(1) \sum_{|\alpha| \leq m } |\xi^{\alpha}|^2 \leq (1+ |\xi|^2)^m \leq c_m \sum_{|\alpha| \leq m} |\xi^{\alpha}|^2$$
We have $\xi^{\alpha} \widehat{u}(\xi)=\widehat{D^{\alpha} u}$, so $D^{\alpha} u \in L^2(\mathbb{R}^n) \Rightarrow \xi^{\alpha} \widehat{u}(\xi) \in L^2(\mathbb{R}^n)$
From $(1)$ we have that the two metrics $(\star)$ and $||u||_m^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}$ are equivalent.
Claim: each Cauchy sequence $(\phi_j)_{j=1}^{\infty} \in C_C^{\infty}(\mathbb{R}^n)$ as for the metric $||\cdot||_m$ converges to an element of $H_m(\mathbb{R}^n)$.
We have that $\phi_j$ is Cauchy in $L^2(\mathbb{R}^n)$, $D^{\alpha}\phi_j$ is Cauchy in $L^2(\mathbb{R}^n)$.
$||\phi_j||_m^2=\sum_{|\alpha| \leq m} |D^{\alpha} \phi_j|^2 dx$
Let $u$ the $L^2$-limit of $\phi_j$ and $u^{\alpha}$ is the $L^2$-limit of $D^{\alpha}\phi_j$ , then
$$\int_{\mathbb{R}^n} \psi D^{\alpha} \phi_j=(-1)^{|\alpha|} \int_{\mathbb{R}^n} D^{\alpha} \psi \phi_j dx, \psi \in C_C^{\infty}(\mathbb{R}^n)$$
$$\Rightarrow \int_{\mathbb{R}^n} \psi u^{\alpha} dx=(-1)^{|\alpha|} \int D^{\alpha}{\psi} u dx \Rightarrow u^{\alpha}=D^{\alpha} u$$
Furthermore, each function $u \in H_m(\mathbb{R}^n)$ is the limit of a sequence $(\phi_j)_{j=1}^{\infty}$.
We define $u_{\epsilon}(x)= \int_{\mathbb{R}^n} \psi_{\epsilon}(x-y) u(y) dy$.
$$||u_{\epsilon}||_{L^2} \leq ||u||_{L^2}, ||D^{\alpha} u_{\epsilon}||_{L^2} \leq ||D^{\alpha} u||_{L^2}$$
$$||D^{\alpha} u_{\epsilon}- D^{\alpha} u||_{L^2(\mathbb{R}^n)} \to 0$$
We consider a cutoff function $\phi \in C_C^{\infty}(\mathbb{R}^n), 0 \leq \phi \leq 1$ and $\phi=1$ for $|x|<1$.
We define $g_{\epsilon}(x)=\phi( \epsilon x) u_{\epsilon}(x)$.
$D^{\alpha} g_{\epsilon}-D^{\alpha} u=D^{\alpha} (\phi(\epsilon x) u_{\epsilon }x)-D^{\alpha} u(x)= \phi(\epsilon x) D^{\alpha} u_{\epsilon}(x)- \phi(\epsilon x) D^{\alpha} u+ \phi(\epsilon x) D^{\alpha}u-D^{\alpha} u+ \epsilon(\dots)=\phi(\epsilon x) (D^{\alpha} u_{\epsilon}(x)-D^{\alpha}u)+(\phi(\epsilon x)-1) D^{\alpha}u+ \epsilon (\dots)$
$\Rightarrow D^{\alpha} g_{\epsilon} \overset{L^2}{\rightarrow} D^{\alpha}u$
Firts of all, how does the relation $(1)$ imply that the metrics $(\star)$ and $||u||_m^2=(2 \pi)^{-n} \int (1+|\xi|^2)^m |\widehat{u}(\xi)|^2 d{\xi}$ are eqivalent?
Secondly, in order to show that $H_m(\mathbb{R}^n)$ is the completion of $C_{C}^{\infty}(\mathbb{R}^n)$, do we need to show that each sequence of $C_C^{\infty}(\mathbb{R}^n)$ converges to an element of $H_m(\mathbb{R}^n)$ and that each element of $H_m(\mathbb{R}^n)$ is the limit of a sequence of $C_C^{\infty}(\mathbb{R}^n)$ ? If so, why?
Why are $\phi_j$ and $D^{\alpha} \phi_j$ Cauchy sequences in $L^2(\mathbb{R}^n)$ ?
In addition, why do we look for the difference $D^{\alpha}g_{\epsilon}-D^{\alpha}u $ ?
And how did we calculate $D^{\alpha}(\phi( \epsilon x) u_{\epsilon}(x))$ ?